Question: Is ${487818}$ divisible by $9$ ?
A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {487818}= &&{4}\cdot100000+ \\&&{8}\cdot10000+ \\&&{7}\cdot1000+ \\&&{8}\cdot100+ \\&&{1}\cdot10+ \\&&{8}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {487818}= &&{4}(99999+1)+ \\&&{8}(9999+1)+ \\&&{7}(999+1)+ \\&&{8}(99+1)+ \\&&{1}(9+1)+ \\&&{8} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {487818}= &&\gray{4\cdot99999}+ \\&&\gray{8\cdot9999}+ \\&&\gray{7\cdot999}+ \\&&\gray{8\cdot99}+ \\&&\gray{1\cdot9}+ \\&& {4}+{8}+{7}+{8}+{1}+{8} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${487818}$ is divisible by $9$ if ${ 4}+{8}+{7}+{8}+{1}+{8}$ is divisible by $9$ Add the digits of ${487818}$ $ {4}+{8}+{7}+{8}+{1}+{8} = {36} $ If ${36}$ is divisible by $9$ , then ${487818}$ must also be divisible by $9$ ${36}$ is divisible by $9$, therefore ${487818}$ must also be divisible by $9$.